Respuesta :
Using the z-distribution, as we are working with a proportion, it is found that:
a) The population proportion is of 0.625.
b) The interval is (0.586, 0.664).
What is a confidence interval of proportions?
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] is the sample proportion.
- z is the critical value.
- n is the sample size.
In this problem, we have a 99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.575.
The sample size and the population proportion are given, respectively, by:
[tex]n = 1040, \pi = \frac{650}{1040} = 0.625[/tex]
Hence, the bounds of the interval are given by:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.625 - 2.575\sqrt{\frac{0.625(0.375)}{1040}} = 0.586[/tex]
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.625 + 2.575\sqrt{\frac{0.625(0.375)}{1040}} = 0.664[/tex]
More can be learned about the z-distribution at https://brainly.com/question/25890103
